Integrand size = 28, antiderivative size = 175 \[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^3 \, dx=\frac {26 a^3 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{21 d}+\frac {26 i a^3 (e \sec (c+d x))^{5/2}}{35 d}+\frac {26 a^3 e (e \sec (c+d x))^{3/2} \sin (c+d x)}{21 d}+\frac {2 i a (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^2}{9 d}+\frac {26 i (e \sec (c+d x))^{5/2} \left (a^3+i a^3 \tan (c+d x)\right )}{63 d} \]
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Time = 0.25 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3579, 3567, 3853, 3856, 2720} \[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^3 \, dx=\frac {26 a^3 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{21 d}+\frac {26 i a^3 (e \sec (c+d x))^{5/2}}{35 d}+\frac {26 a^3 e \sin (c+d x) (e \sec (c+d x))^{3/2}}{21 d}+\frac {26 i \left (a^3+i a^3 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}{63 d}+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{5/2}}{9 d} \]
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Rule 2720
Rule 3567
Rule 3579
Rule 3853
Rule 3856
Rubi steps \begin{align*} \text {integral}& = \frac {2 i a (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^2}{9 d}+\frac {1}{9} (13 a) \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^2 \, dx \\ & = \frac {2 i a (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^2}{9 d}+\frac {26 i (e \sec (c+d x))^{5/2} \left (a^3+i a^3 \tan (c+d x)\right )}{63 d}+\frac {1}{7} \left (13 a^2\right ) \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x)) \, dx \\ & = \frac {26 i a^3 (e \sec (c+d x))^{5/2}}{35 d}+\frac {2 i a (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^2}{9 d}+\frac {26 i (e \sec (c+d x))^{5/2} \left (a^3+i a^3 \tan (c+d x)\right )}{63 d}+\frac {1}{7} \left (13 a^3\right ) \int (e \sec (c+d x))^{5/2} \, dx \\ & = \frac {26 i a^3 (e \sec (c+d x))^{5/2}}{35 d}+\frac {26 a^3 e (e \sec (c+d x))^{3/2} \sin (c+d x)}{21 d}+\frac {2 i a (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^2}{9 d}+\frac {26 i (e \sec (c+d x))^{5/2} \left (a^3+i a^3 \tan (c+d x)\right )}{63 d}+\frac {1}{21} \left (13 a^3 e^2\right ) \int \sqrt {e \sec (c+d x)} \, dx \\ & = \frac {26 i a^3 (e \sec (c+d x))^{5/2}}{35 d}+\frac {26 a^3 e (e \sec (c+d x))^{3/2} \sin (c+d x)}{21 d}+\frac {2 i a (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^2}{9 d}+\frac {26 i (e \sec (c+d x))^{5/2} \left (a^3+i a^3 \tan (c+d x)\right )}{63 d}+\frac {1}{21} \left (13 a^3 e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {26 a^3 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{21 d}+\frac {26 i a^3 (e \sec (c+d x))^{5/2}}{35 d}+\frac {26 a^3 e (e \sec (c+d x))^{3/2} \sin (c+d x)}{21 d}+\frac {2 i a (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^2}{9 d}+\frac {26 i (e \sec (c+d x))^{5/2} \left (a^3+i a^3 \tan (c+d x)\right )}{63 d} \\ \end{align*}
Time = 2.74 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.51 \[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^3 \, dx=\frac {a^3 \sec ^2(c+d x) (e \sec (c+d x))^{5/2} \left (728 i+1008 i \cos (2 (c+d x))+1560 \cos ^{\frac {9}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-150 \sin (2 (c+d x))+195 \sin (4 (c+d x))\right )}{1260 d} \]
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Time = 7.50 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.05
\[-\frac {2 i e^{2} a^{3} \sqrt {e \sec \left (d x +c \right )}\, \left (195 F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+195 F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+195 i \tan \left (d x +c \right )-252 \left (\sec ^{2}\left (d x +c \right )\right )-135 i \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )+35 \left (\sec ^{4}\left (d x +c \right )\right )\right )}{315 d}\]
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.47 \[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^3 \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (195 i \, a^{3} e^{2} e^{\left (8 i \, d x + 8 i \, c\right )} - 1158 i \, a^{3} e^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 1456 i \, a^{3} e^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 858 i \, a^{3} e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 195 i \, a^{3} e^{2}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 195 \, \sqrt {2} {\left (i \, a^{3} e^{2} e^{\left (8 i \, d x + 8 i \, c\right )} + 4 i \, a^{3} e^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 i \, a^{3} e^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 4 i \, a^{3} e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{3} e^{2}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )}}{315 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
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\[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^3 \, dx=- i a^{3} \left (\int i \left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}\, dx + \int \left (- 3 \left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}} \tan {\left (c + d x \right )}\right )\, dx + \int \left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}} \tan ^{3}{\left (c + d x \right )}\, dx + \int \left (- 3 i \left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}} \tan ^{2}{\left (c + d x \right )}\right )\, dx\right ) \]
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\[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^3 \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} \,d x } \]
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\[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^3 \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} \,d x } \]
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Timed out. \[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^3 \, dx=\int {\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3 \,d x \]
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